Forgetting my Integration

[MrJoe2000]

I have to solve the problem:

Int(a + bx)^1/2

I'm getting [2*(a + bx)^3/2]/3 + C

the book is getting [2*(a + bx)^3/2]/3b + C

Where are they getting the "b" term in the denominator from?

 

[srmichael]

\(\displaystyle u=a+bx\)

\(\displaystyle du=(b)(dx)\)

\(\displaystyle dx=\frac{du}{b}\)

Now you have your "b" in the denominator

 

[MrJoe2000]

So when you differentiate the f(x) you found by taking the integral, how do you cancel out the demonimator?

 

[renegade05]

So when you differentiate the f(x) you found by taking the integral, how do you cancel out the demonimator?

When you differentiate \(\displaystyle (a + bx)^{3/2}\) you will use the chain rule.

So \(\displaystyle \frac{d}{dx}((a + bx)^{3/2}) = \frac{3}{2}(a + bx)^{1/2}b\) The b's will cancel out.